Aball is thrown downward from the top of a building with an initial speed of 25m/s. it strikes the ground after 2.0s. how high is the building?

For this case we have an equation of the form:
 h (t) = (1/2) * (a) * (t ^ 2) + vo * t + h0
 Where,
 a: acceleration
 vo: initial speed
 h0: initial height
 By the time the ball hits the ground we have:
 h (t) = 0
 Substituting values:
 0 = - (1/2) * (9.8) * (2 ^ 2) - 25 * 2 + h0
 h0 = (1/2) * (9.8) * (2 ^ 2) + 25 * 2
 h0 = 69.6 m
 
 the building is 69.6 m high